Self-induction of Wires. 431 



from which we see that 



X=J.C/i), Y=K„(/*), 

 where 



/=(-E «S p)*. 



But, owing to the infinite conductivity at the z = end of the 

 line, making K (/z) = go there, we shall only be concerned 

 with the J function, that is, on the left side of the impressed 

 force, in the first place. Since V is made permanently zero 

 at z=0, the terminal condition there is nugatory. So 



™ = J (f z )> and w = J o (/*0+2iK o (/s) ; 



"=(fl&P)Ji(fi>)> a * d u=(fJSp){J 1 (fz) + q 1 K 1 (fi)}; 



on the left and right sides of an impressed force, say at z = z 2 . 

 The value of q i} got from the Y=Z 1 C condition at z=l } is 



_ (/ys oP )j 1 (/o-z 1 j (/0 nfin s 



? 1 -Z 1 K„(/0-(///8 0i >)K 1 (/0- • * • libUo; 

 We have also 



XY'— YX' f 1 



Wp = Wp ( J i K o- JoKiX/s) = g-^ ; (155a) 



and the C solution (166) becomes 



G=t(-p$)-V a (fz) £eJ (fz)dz. (l-*>), (166a) 



where cj)= — q 1 /$ p, and q l is given by (160a). 



If we short-circuit at z=l, making Z^O, we introduce 

 peculiarities connected with the presence of the series of jo's 

 belonging to/=0. The expression of q 1 is then, by (160a), 

 q 1 =—J 1 (y7)/K 1 (//). It seems rather singular that we should 

 have anything to do with the K x function, seeing that C and 

 Y are expanded in series of the J and J\ functions. But on 

 performing the differentiation of <f) with respect to p it turns 

 out to be all right, the denominator in (166a) becoming 



in general ; whilst in the /=0 case, which makes <p = ^B JQ ,, l 2 y 

 we have 



The value of </> when p = in it is, by inspection of the 

 expansions of J 1 and Ki, simply £iy 2 , the steady-flow resist- 



